The sides of the Harriss rectangle assuming the basic square has side 1:

{A, B} = N[{a, b} /. Solve[a/b == b/(a - 1) == 1/(b - 1), {a, b}][[1]]]

{2.32472, 1.75488}

A/B

1.32472

Recursive code. eps is a discretization to stop the iteration. The second argument tells whether we want to show the discretization process or not. p and q in the recursive function are the corners of the next rectangle to be subdivided:

HarrissSpiral[eps_, squares_] := 
  Graphics[{EdgeForm[Black], 
    HarrissSpiralStep[{0, 0}, {A, B}, eps, squares]}, 
   AspectRatio -> Automatic];
HarrissSpiralStep[p_, q_, eps_, squares_] := 
 If[EuclideanDistance[p, q] > eps,
  With[{
    r = p + (A - 1)/
        Sqrt[A^2 + B^2] RotationMatrix[-ArcTan[A, B]].(q - p),
    s = p + B/Sqrt[A^2 + B^2] RotationMatrix[ArcTan[B, A]].(q - p),
    t = p + Sqrt[1 + (A - 1)^2]/Sqrt[A^2 + B^2] (q - p)},
   {If[squares,
     {
      Black, Arrow[{p, q}],
      White, Rectangle[p, q],
      Blue, Rectangle[r, s],
      Orange, Rectangle[t, q]
      },
     {}
     ],
    Red, AbsoluteThickness[5 Norm[s - p]], 
    Circle[p + RotationMatrix[-Pi/4].(s - p)/Sqrt[2], 
     Norm[s - p]/Sqrt[2], ArcTan @@ (s - p) + {Pi/4, 3 Pi/4}],
    HarrissSpiralStep[r, s, eps, squares],
    HarrissSpiralStep[t, q, eps, squares]
    }],
  {}
  ]

Only a few iterations:

HarrissSpiral[0.8, True]

More iterations:

HarrissSpiral[0.1, True]

Show only the quarter circles:

HarrissSpiral[0.02, False]

To understand the idea, the key is to follow the arrows, which determine the orientation of the rectangles. There is a blue or orange rectangle and a quarter circle per arrow. Some arrows are overlapped. All arrows are parallel or perpendicular.

Graphics[{EdgeForm[Black], Opacity[0.3],  HarrissSpiralStep[{0, 0}, {A, B}, 0.7, True]}, 
     AspectRatio -> Automatic]

Attachments:

Thanks Sam.

1) Yes, it would be possible to use AffineTransform here to do the rescalings+rotations+shifts I do manually above. And the code would look simpler and neater. I did it this way because then I can reuse the points r, s, t to do tests and variations during the iteration.

2) As far as I see, the construction given above is fine tuned to place the rectangles in particular places, so that the quarter circles match, producing a continuous result. Changing the values of A and B would break that. As the author shows, it is possible to produce other spirals of this type by careful choice of how to replicate the rectangles (that means changing the affine transforms in the code). It would be great to find some continuous parameters that allow interpolation among (at least some of) those alternative spirals. Perhaps one would have to relax the rule of using only quarter circles for the curves.

Unfortunately no, I would like to know how to build those myself. Picture is simply form the website linked by Vitaliy.